Two baseball players bat first and second in the lineup. The first batter has an on-base percentage of 0.23. The second batter has an on-percentage of 0.38 if someone is on base, but only 0.26 if the bases are empty. At the start of the game, what is the probability that neither player gets on base?

Answers

Answer 1

Answer:

The probability that neither player gets on base is 0.4824

Step-by-step explanation:

1. Both players get to base. Just multiply the two probabilities together:

= (probability first batter gets on base) x (probability second batter gets on base, if the first batter gets on base)

= 0.23 x 0.38

= 0.0874

2. One player gets to base. The formula here is P(A+B) =P(A) + P(B) - P(A) x P(B)

= (probability first batter gets on base) + (probability second batter gets on base, if the first batter does not) - (0.23 x 0.26)

= 0.23 + 0.26 - (0.23 x 0.26)

= 0.49 - 0.0598

= 0.4302

3. Neither player gets to base = 1 - addition of the previous two cases.

= 1 - (0.0874 + 0.4302)

= 1 - 0.5176

= 0.4824


Related Questions

A bag of walnuts sells for$1.95 Estimate the cost of 8 bags of walnut

Answers

Answer:

$15.6

Step-by-step explanation:

one bag= $1.95

8 bags = 8×1.95=15.6

so answer is $15.6.

help, will give brainliest

Answers

Answer:

10.0

Step-by-step explanation:

4.0+8.0=12.0-2.0=10.0

40% of students in a school wore blue on a spirit Day. If two
students are randomly selected, what is the probability that both
students will not be wearing blue?

Answers

The probability that both students will not wear blue is 0.36, obtained by multiplying the probability of one student not wearing blue, 0.60, by itself.

To find the probability that both students will not be wearing blue, we first need to determine the probability that one student is not wearing blue, and then multiply that probability by itself for two students.

Step 1: Calculate the probability that one student is not wearing blue.

Given that 40% of students wore blue, the probability that one student is not wearing blue is 1 - 0.40 = 0.60.

Step 2: Multiply the probability for one student by itself for two students.

The probability that both students will not be wearing blue is (0.60) ×(0.60) = 0.36.

So, the probability that both students will not be wearing blue is 0.36.

3•(7+10)=G+30 use the distributive property to solve

Answers

Answer: G=21

Step-by-step explanation:

Solve for G by simplifying both sides of the equation, then isolating the variable.

What is the volume, in cubic in, of a rectangular prism with a height of 5in, a width of 12in, and a length of 14in?

Answers

Answer:

840

Step-by-step explanation:

the volume of a rectangular prism is the base(width)height

When the driver applies the brakes of a small-size truck traveling 10 mph, it skids 5 ft before stopping. How far will the truck skid if it is traveling 55 mph when the brakes are applied

Answers

Answer:

The truck will skid for 151 ft before stopping when the brakes are applied

Step-by-step explanation:

From the equations of motion, we will use

[tex]v^{2} = u^{2}-2aS[/tex]

We have to make sure that the parameters we are working with are in the same unit of length. Here, we will be converting from ft to miles

When the truck is travelling at 10 mph.

S = distance the truck skids = [tex]\frac{5ft}{5286ft/mile}= 0.00094697miles[/tex]

Final velocity of truck, v = 0 m/s (this is because the truck decelerates to a halt)

Initial velocity of truck u = 10 mph

Hence, we have

[tex]0^{2}=10^{2}-2a\times 0.00094697[/tex]

[tex]a= 52799.9miles/hr^{2}[/tex]

This is the deceleration of the truck

We will work based on the assumption that the car decelerates at the same rate each time the brakes are fully applied.

When the truck is travelling at u= 55 mph.

We will need to use the deceleration of the car to find the distance traveled when it skids.

[tex]0^{2}=55^{2}-2\times52799.98\times S[/tex]

[tex]S= 0.0286 miles\approx 151 ft[/tex]

∴The car skids for about 151 ft when it is travelling at 55 mph and the brakes are applied.

Drag each tile to the correct cell in the table.
High school students were surveyed about which math and
science topics they preferred. They were asked to choose
algebra or geometry, and biology or chemistry. The results are
shown in the frequency table below.
QUI 73%DC 100%
C
70%
30%
30%
100%
100%
28%
72%
Use this table to create a relative frequency table by row.
27%
Algebra
Geometry
Total
Algebra
Geometry
Total
Biology
67
92
Biology
Chemistry
46
66
Chemistry
Total
113
45
158
Total

Answers

Answer:

Step-by-step explanation:

Table(2) shows the relative frequency opted from the table(1).

What is the frequency?

It is defined as the number of waves that crosses a fixed point in one second known as frequency. The unit of frequency is per second.

We have a table in which data has shown:

To find the relative frequency:

For the first cell:

[tex]=\frac{67}{92} \times100 \approx 73\%[/tex]

For the second cell:

= 100 - 73 ⇒ 27%

For the third cell:  

[tex]\rm = \frac{46}{66} \times 100 \approx 70\%[/tex]

For the fourth cell:

= 100 - 70 = 30%

Thus, table(2) shows the relative frequency opted from table(1).

Learn more about the frequency here:

brainly.com/question/27063800

Suppose 40% of DC area adults have traveled outside of the United States. Nardole wants to know if his customers are atypical in this respect. He surveys 40 customers and finds 60% have traveled outside of the U.S. Is this result a statistically significant difference?

Answers

Answer:

We conclude that % of DC area adults who have traveled outside of the United States is different from 40%.

Step-by-step explanation:

We are given that 40% of DC area adults have traveled outside of the United States. Nardole wants to know if his customers are typical in this respect. He surveys 40 customers and finds 60% have traveled outside of the U.S.

We have to test is this result a statistically significant difference.

Let p = % of DC area adults who have traveled outside of the United States

SO, Null Hypothesis, [tex]H_0[/tex] : p = 40%  {means that 40% of DC area adults have traveled outside of the United States}

Alternate Hypothesis, [tex]H_a[/tex] : p [tex]\neq[/tex] 40%  {means that % of DC area adults who have traveled outside of the United States is different from 40%}

The test statistics that will be used here is One-sample z proportion statistics;

                  T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = % of customers who have traveled outside of the United States

                  in a survey of 40 customers = 60%

          n  = sample of customers = 40

So, test statistics = [tex]\frac{0.60-0.40}{\sqrt{\frac{0.60(1-0.60)}{40} } }[/tex]

                             = 2.582

Since in the question we are not given with the significance level so we assume it to be 5%. So, at 0.05 level of significance, the z table gives critical value of 1.96 for two-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that % of DC area adults who have traveled outside of the United States is different from 40%.

three less than the product of 4 and the number is equal to 5

Answers

Answer:

4n-3=5

Step-by-step explanation:

If you read it step by step the product of and a number is basically 4* any variable.( In this case I use n.) Next the beginning part is 3 less than the product part so it is 4n-3. And finally the whole equation is equal to 5 so itis 4n-3=5.

The average sales per customer at a home improvement store during the past year is $75 with a standard deviation of $12. The probability that the average sales per customer from a sample of 36 customers, taken at random from this population, exceeds $78 is:

Answers

Answer:

0.0668

Step-by-step explanation:

Assuming the distribution is normally distributed with a mean of $75,

with a standard deviation of $12.

We can find the z-score of 78 using;

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

[tex]\implies z=\frac{78-75}{\frac{12}{36} } =1.5[/tex]

Using our normal distribution table, we obtain the area that corresponds to 0.25 to be 0.9332

This is the area corresponding to the probability that, the average is less or equal to 78.

Subtract from 1 to get the complement.

P(x>78)=1-0.9332=0.0668

The probability that the average sales per customer from a sample of 36 customers, taken at random from this population, exceeds $78 is 0.0668.

Calculation of the probability:

Since The average sales per customer at a home improvement store during the past year is $75 with a standard deviation of $12.

Here  we need to find out the z score

= [tex]78-75\div 12\div 36[/tex]

= 1.5

Here we considered normal distribution table, we obtain the area that corresponds to 0.25 to be 0.9332

So,  the average is less or equal to 78.

Now

Subtract from 1 to get the complement.

So,

P(x>78)=1-0.9332

=0.0668

Learn more about probability here: https://brainly.com/question/24613748

Please help
What is the slope of a line that is parallel to the line shown?

Answers

Answer:

m=2/3

Step-by-step explanation:

your answer would be A.

2 up and 3 over. So it will be A) 2/3

A school has two kindergarten classes. There are 21 children in Ms. Toodle's kindergarten class. Of these, 17 are "pre-readers" children on the verge of reading. There are 19 children in Mr. Grimace's kindergarten class. Of these, 13 are pre-readers. Using the plus four confidence interval method, a 90% confidence interval for the difference in proportions of children in these classes that are pre-readers is â0.104 to 0.336.

Which of the following statements is correct?

A) This confidence interval is not reliable because the samples are so small.
B)This confidence interval is of no use because it contains 0, the value of no difference between classes.
C)This confidence interval is reasonable because the sample sizes are both at least 5.
D) This confidence interval is not reliable because these samples cannot be viewed as simple random samples taken from a larger population.

Answers

Answer:

Answer : D

Step-by-step explanation:

A school has two kindergarten classes. There are 21 children in Ms. Toodle's kindergarten class. Of these, 17 are "pre-readers" children on the verge of reading. There are 19 children in Mr. Grimace's kindergarten class. Of these, 13 are pre-readers. Using the plus four confidence interval method, a 90% confidence interval for the difference in proportions of children in these classes that are pre-readers is â0.104 to 0.336.  

Which of the following statements is correct?

A) This confidence interval is not reliable because the samples are so small.  

B)This confidence interval is of no use because it contains 0, the value of no difference between classes.  

C)This confidence interval is reasonable because the sample sizes are both at least 5.  

D) This confidence interval is not reliable because these samples cannot be viewed as simple random samples taken from a larger population.

The Answer is D - This confidence interval is not reliable because these samples cannot be viewed as simple random samples taken from a larger population.

In this setup, all the students are already involved in the data. This is not a sample from a larger population, but probably, the population itself.

Daniel makes 16 more muffins than kris.
Daniel makes 34 muffins. How many muffins does kris make?
Solve the equation problem choose yes or no

Answers

Kris makes 18 muffins

Answer the question.
Sophie works as a computer programmer. she earns $28 per hour. If Sophie works 10 hours. how much money will she earn?

Answers

$280

10 * 28 = 280 hope this helps :)

What is the median of the following data set?

{6, 3, 9, 1, 7}

3
6
8
9
I know the answer just seeing what you guys know and giving points:)

Answers

Answer:

The median of the following set of data is 9 since the question is implying, which is the center of the data distribution.

Step-by-step explanation:

Answer: 6
The median is the number in the middle after putting the points in ascending order.

What is the probability that a domestic airfare is $560 or more (to 4 decimals)? b. What is the probability that a domestic airfare is $260 or less (to 4 decimals)? c. What if the probability that a domestic airfare is between $320 and $500 (to 4 decimals)? d. What is the cost for the 3% highest domestic airfares? (rounded to nearest dollar)

Answers

Final answer:

Without sufficient data, we cannot calculate the probabilities or the cost for the 3% highest domestic airfares. The calculations require details about the total number of observed airfares and the number that falls in the specified price ranges.

Explanation:

I regret to inform you that I cannot provide a factual answer to your question regarding domestic airfares as you have not provided sufficient data. Probability depends on the sample space and given conditions which are not provided in your question. For instance:

For the probability of a domestic airfare being $560 or more, we need to know the total number of observed airfares and the number that were $560 or more. The same applies to a domestic airfare of $260 or less (to 4 decimals). The total fares and the ones falling under this range are required. The probability of a fare being between $320 and $500 is calculated from the total fares within this range divided by the total fares observed. For the 3% highest fares, we need to know the distribution of airfares to calculate this value.

Learn more about Probability here:

https://brainly.com/question/22962752

#SPJ11

g a. A 99% confidence level and a sample of 24 observations. 2.807 2.807 Correct b. A 90% confidence level and a sample of 24 observations. 1.714 1.714 Correct c. A 99% confidence level and a sample of 20 observations. 2.845 2.845 Incorrect d. A 90% confidence level and a sample of 20 observations.

Answers

Answer: b. A 90% confidence level and a sample of 24 observations. 1.714 1.714 Correct

Step-by-step explanation:

Victor has $40 in a savings account. The interest rate is 5%, compounded annually.
To the nearest cent, how much interest will he earn in 3 years?

Answers

Answer:

$6.31

Step-by-step explanation:

We are going to use the compound simple interest formula for this problem:

[tex]A=P(1+\frac{r}{n} )^{nt}[/tex]

P = initial balance

r = interest rate (decimal)

n = number of times compounded annually

t = time

Our first step is to change 5% into its decimal form:

5% -> [tex]\frac{5}{100}[/tex] -> 0.05

Next, plug in the values:

[tex]A=40(1+\frac{0.05}{1})^{1(3)}[/tex]

[tex]A=46.31[/tex]

Lastly, subtract 40 (our original value) from 46.31:

[tex]46.31-40=6.31[/tex]

Victor earned $6.31 in interest after 3 years.

If m ≤ f(x) ≤ M for a ≤ x ≤ b, where m is the absolute minimum and M is the absolute maximum of f on the interval [a, b], then m(b − a) ≤ b f(x) dx a ≤ M(b − a). Use this property to estimate the value of the integral. π/6 5 tan(2x) dx π/8

Answers

Answer:

The final integration in the given limits will be 89.876

n monitoring lead in the air after the explosion at the battery factory, it is found that the amounts of lead over a 6 day period had a standard error of 1.91. Find the margin of error that corresponds to a 95% confidence interval. (Round to 2 decimal places)

Answers

Answer:

434.98

Step-by-step explanation:

A fruit stand sells 6 oranges for $3.00 and 3 grapefruit for $2.40. Sherry buys 10 oranges and 11 grapefruit. How much does Sherry spend on the fruit?

Answers

Answer:

Step-by-step explanation:

FIRST FOR ORANGES SHE BOUGHT 10 SO 10 TIMES 3 IS 30 AND THEN 11 TIMES 2.40 IS 26.40. 30 PLUS 26.40 IS 56.40 THATS YOUR AWNSER

Which expression is equivalent to 15x – 2(3x + 6)?

Answers

Answer: 3(3x−4) is the answer

Answer:

3(3x-4) is the answer

Step-by-step explanation:

1. $10 coupon on a $50.00 dinner

Answers

Answer: It would make the meal $40.00 and with tax it would be $42.80

Step-by-step explanation: 50 - 10 = 40

                                              40 / 0.07 = 2.8

                                              40 + 2.8 = 42.8

Answer:

Step-by-step explanation:

The screening process for detecting a rare disease is not perfect. Researchers have developed a blood test that is considered fairly reliable. It gives a positive reaction in 94.5% of the people who have that disease. However, it erroneously gives a positive reaction in 1.5% of the people who do not have the disease. Consider the null hypothesis "the individual does not have the disease" to answer the following questions.
a. What is the probability of a Type I error?
b. What is the probability of a Type II error?

Answers

Answer:

a) Type 1 Error: 1.5%

b) Type 2 Error: 5.5%

Step-by-step explanation:

Probability of positive reaction when infact the person has disease = 94.5%

This means, the probability of negative reaction when infact the person has disease = 100- 94.5% = 5.5%

Probability of positive reaction when the person does not have the disease = 1.5%

This means,

Probability of negative reaction when the person does not have disease = 100% - 1.5% = 98.5%

Our Null Hypothesis is:

"The individuals does not have the disease"

Part a) Probability of Type 1 Error:

Type 1 error is defined as: Rejecting the null hypothesis when infact it is true. Therefore, in this case the Type 1 error will be:

Saying that the individual have the disease(positive reaction) when infact the individual does not have the disease. This means giving a positive reaction when the person does not have the disease.

From the above data, we can see that the probability of this event is 1.5%. Therefore, the probability of Type 1 error is 1.5%

Part b) Probability of Type 2 Error:

Type 2 error is defined as: Accepting the null hypothesis when infact it is false. Therefore, in this case the Type 2 error will be:

Saying that the individual does not have the disease(negative reaction) when infact the individual have the disease.

From the above data we can see that the probability of this event is 5.5%. Therefore, the probability of Type 2 error is 5.5%

Which of the following equations represents a hyperbola with a vertical transverse axis and foci at (0, −10) and (0, 10)?

Answers

Answer: A

Step-by-step explanation:

Answer:

the first opition is the correct answer

-2/5x - 5 = -25

solve X QUICKKKKLY

Answers

Answer:

50

Step-by-step explanation:

add 5 to -25 to get -2/5x by itself.

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that join them. Find the center of mass of the lamina if the density at any point is inversely proportional to its distance from the origin.

Answers

The center of mass of the lamina is located at the point (0, (39/12) (1 / ln(4))).

To find the center of mass of the given lamina, we need to calculate the moment about the x-axis and the y-axis, and then divide them by the total mass of the lamina.

Given information:

- The boundary of the lamina consists of the semicircles y = sqrt(1 - x^2) and y = sqrt(16 - x^2), and the portions of the x-axis that join them.

- The density at any point is inversely proportional to its distance from the origin.

Find the total mass of the lamina.

Let the density function be [tex]\[ \rho(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \][/tex], where k is a constant.

The total mass, M, is given by the double integral of the density function over the region of the lamina.

M = ∫∫ ρ(x, y) dA

To evaluate this integral, we need to express the lamina in polar coordinates.

The semicircles can be represented as:

0 ≤ r ≤ 1, 0 ≤ θ ≤ π

0 ≤ r ≤ 4, π ≤ θ ≤ 2π

The total mass can be calculated as:

[tex]\[ M = \int_{0}^{\pi} \int_{0}^{1} \frac{k}{r} r \, dr \, d\theta + \int_{\pi}^{2\pi} \int_{0}^{4} \frac{k}{r} r \, dr \, d\theta \]\[ M = k \left( \pi \ln(1) + 2\pi \ln(4) \right) \]\[ M = 2\pi k \ln(4) \][/tex]

Calculate the moment about the x-axis.

The moment about the x-axis, Mx, is given by:

[tex]\[ M_x = \int_{0}^{\pi} \int_{0}^{1} \frac{k}{r} r^2 \sin(\theta) \, dr \, d\theta + \int_{\pi}^{2\pi} \int_{0}^{4} \frac{k}{r} r^2 \sin(\theta) \, dr \, d\theta \]\[ M_x = k \left( \frac{\pi}{2} + \frac{32\pi}{3} \right) \]\[ M_x = \frac{39\pi}{6} k \][/tex]

Calculate the moment about the y-axis.

The moment about the y-axis, My, is given by:

My = ∫∫ x ρ(x, y) dA

In polar coordinates:

[tex]\[ M_y = \int_{0}^{\pi} \int_{0}^{1} \frac{k}{r} r^2 \cos(\theta) \, dr \, d\theta + \int_{\pi}^{2\pi} \int_{0}^{4} \frac{k}{r} r^2 \cos(\theta) \, dr \, d\theta \][/tex]

My = 0 (due to symmetry)

Find the coordinates of the center of mass.

The coordinates of the center of mass (x_cm, y_cm) are given by:

x_cm = My / M

y_cm = Mx / M

Substituting the values, we get:

x_cm = 0 / (2πk ln(4)) = 0

y_cm = (39π/6) k / (2πk ln(4)) = (39/12) (1 / ln(4))

Therefore, the center of mass of the lamina is located at the point (0, (39/12) (1 / ln(4))).

Complete question:

The boundary of a lamina consists of the semicircles y=sqrt(1 − x^2) and y= sqrt(16 − x^2) together with the portions of the x-axis that join them. Find the center of mass of the lamina if the density at any point is inversely proportional to its distance from the origin.

The center of mass of the lamina is at the origin (0, 0)

To find the center of mass of the lamina, we first need to find the mass and the moments about the  x- and  y-axes.

The mass  M of the lamina can be calculated by integrating the density function over the lamina. Since the density at any point is inversely proportional to its distance from the origin, we can express the density[tex]\( \delta \) as \( \delta(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \)[/tex], where  k is a constant.

Let's denote [tex]\( \delta(x, y) \) as \( \frac{k}{\sqrt{x^2 + y^2}} \)[/tex]. Then the mass M  is given by the double integral of [tex]\( \delta(x, y) \)[/tex] over the region  R bounded by the semicircles and the portions of the x-axis:

[tex]\[ M = \iint_R \delta(x, y) \, dA \][/tex]

Where  dA  represents the differential area element.

To find the moments about the  x- and  y -axes, we calculate:

[tex]\[ M_x = \iint_R y \delta(x, y) \, dA \]\[ M_y = \iint_R x \delta(x, y) \, dA \][/tex]

Then, the coordinates [tex]\( (\bar{x}, \bar{y}) \)[/tex] of the center of mass are given by:

[tex]\[ \bar{x} = \frac{M_y}{M} \]\[ \bar{y} = \frac{M_x}{M} \][/tex]

Now, let's proceed to find [tex]\( M \), \( M_x \), and \( M_y \)[/tex]

First, let's express the density [tex]\( \delta(x, y) \)[/tex] in terms of k:

[tex]\[ \delta(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \][/tex]

Now, we'll find the mass  M by integrating [tex]\( \delta(x, y) \)[/tex] over the region  R :

[tex]\[ M = \iint_R \frac{k}{\sqrt{x^2 + y^2}} \, dA \][/tex]

Since the region  R  is symmetric about the  x-axis, we can integrate over the upper half and double the result:

[tex]\[ M = 2 \iint_{R_1} \frac{k}{\sqrt{x^2 + y^2}} \, dA \][/tex]

Now, we'll switch to polar coordinates [tex]\( (r, \theta) \)[/tex]. In polar coordinates, the region [tex]\( R_1 \)[/tex]is described by [tex]\( 0 \leq \theta \leq \pi \) and \( 1 \leq r \leq 4 \).[/tex]

So, the integral becomes:

[tex]\[ M = 2 \int_{0}^{\pi} \int_{1}^{4} \frac{k}{r} \cdot r \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} \int_{1}^{4} 1 \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} (4 - 1) \, d\theta \]\[ = 2k \int_{0}^{\pi} 3 \, d\theta \]\[ = 6k \pi \][/tex]

For [tex]\( M_x \)[/tex], we integrate [tex]\( x \delta(x, y) \)[/tex] over the region R :

[tex]\[ M_x = \iint_R x \cdot \frac{k}{\sqrt{x^2 + y^2}} \, dA \]\[ = 2 \int_{0}^{\pi} \int_{1}^{4} r \cos(\theta) \cdot \frac{k}{r} \cdot r \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} \int_{1}^{4} \cos(\theta) \cdot r \, dr \, d\theta \]\[ = 2k \int_{0}^{\pi} \left[ \frac{1}{2} r^2 \cos(\theta) \right]_{1}^{4} \, d\theta \][/tex]

[tex]\[ = 2k \int_{0}^{\pi} \left( 8 \cos(\theta) - \frac{1}{2} \cos(\theta) \right) \, d\theta \]\[ = 2k \int_{0}^{\pi} \left( \frac{15}{2} \cos(\theta) \right) \, d\theta \]\[ = 2k \left[ \frac{15}{2} \sin(\theta) \right]_{0}^{\pi} \]\[ = 2k \cdot 0 \]\[ = 0 \][/tex]

Now, for[tex]\( M_y \)[/tex], we integrate [tex]\( y \delta(x, y) \)[/tex]over the region  R :

[tex]\[ M_y = \iint_R y \cdot \frac{k}{\sqrt{x^2 + y^2}} \, dA \\\[ = 2 \int_{0}^{\pi} \int_{1}^{4} r \sin(\theta) \cdot \frac{k}{r} \cdot r \, dr \, d\theta \\\[ = 2k \int_{0}^{\pi} \int_{1}^{4} \sin(\theta) \cdot r \, dr \, d\theta \\\[ = 2k \int_{0}^{\pi} \left[ \frac{1}{2} r^2 \sin(\theta) \right]_{1}^{4} \, d\theta \\[/tex]

[tex]\[ = 2k \int_{0}^{\pi} \left( 8 \sin(\theta) - \frac{1}{2} \sin(\theta) \right) \, d\theta \\\[ = 2k \int_{0}^{\pi} \left( \frac{15}{2} \sin(\theta) \right) \, d\theta \\\[ = 2k \left[ -\frac{15}{2} \cos(\theta) \right]_{0}^{\pi} \\\[ = 2k \cdot 0 \\\[ = 0 \][/tex]

Now, we have [tex]\( M = 6k \pi \), \( M_x = 0 \), and \( M_y = 0 \).[/tex]

Finally, we can find the coordinates of the center of mass [tex]\( (\bar{x}, \bar{y}) \):[/tex]

[tex]\[ \bar{x} = \frac{M_y}{M} = \frac{0}{6k \pi} = 0 \]\[ \bar{y} = \frac{M_x}{M} = \frac{0}{6k \pi} = 0 \][/tex]

So, the center of mass of the lamina is at the origin (0, 0) .

Given a family with four ​children, find the probability of the event.

The oldest is a girl and the youngest is a boy​, given that there is at least one boy and at least one girl.

Answers

Answer:

28.57% probability that the oldest is a girl and the youngest is a boy​, given that there is at least one boy and at least one girl.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

These are all the possible outcomes: from youngest to oldest, b is boy and g is girl

b - b - b - g

b - b - g - b

b - b - g - g

b - g - b - b

b - g - b - g

b - g - g - b

b - g - g - g

g - b - b - b

g - b - b - g

g - b - g - b

g - b - g - g

g - g - b - b

g - g - b - g

g - g - g - b

The nuber of total outcomes is 14.

Desired outcomes:

Oldest(last) girl, youngest(first) boy

b - b - b - g

b - b - g - g

b - g - b - g

b - g - g - g

So 4 desired outcomes

Probability:

4/14 = 0.2857

28.57% probability that the oldest is a girl and the youngest is a boy​, given that there is at least one boy and at least one girl.

Final answer:

The probability that a family with four children will have the oldest being a girl and the youngest being a boy, given that there is at least one boy and one girl, is 1/4.

Explanation:

To solve this probability question, we need to consider all possible combinations of children while adhering to the given conditions: the oldest child must be a girl (G), the youngest must be a boy (B), and there must be at least one boy and at least one girl in the family.

The possible genders for the four children can be represented as a sequence of G (girl) and B (boy) like this: G---B. There are two positions in the middle that can be either a boy or a girl. Since each position can be filled independently with either a boy or a girl, there are 2 options for each of the middle children, giving us 2 x 2 = 4 combinations: GBGB, GBBB, GGBB, GGBG.

To calculate the probability of any single one of these combinations occurring, we need to remember that the probability of giving birth to a boy or a girl is equal, which means each event (birth of a child) has a probability of 1/2. Thus, the probability of each combination is (1/2)^4 since there are four independent events (births). However, since we have 4 combinations that meet the criteria, we multiply this probability by 4. So, the probability is 4 * (1/2)^4 = 1/4.

Therefore, the probability that a family with four children will have the oldest being a girl and the youngest being a boy, given that there is at least one boy and at least one girl, is 1/4.

Arc Length and Radians question- please help! Will mark brainliest! Is 20pts!

The answer is shown but please give me an explanation so I can show my work!

Answers

Given:

Given that the radius of the merry - go - round is 5 feet.

The arc length of AB is 4.5 feet.

We need to determine the measure of the minor arc AB.

Measure of the minor arc AB:

The measure of the minor arc AB can be determined using the formula,

[tex]Arc \ length=(\frac{\theta}{360})2 \pi r[/tex]

Substituting arc length = 4.5 and r = 5, we get;

[tex]4.5=(\frac{\theta}{360})2 (3.14)(5)[/tex]

Multiplying the terms, we get;

[tex]4.5=(\frac{\theta}{360})31.4[/tex]

Dividing, we get;

[tex]4.5=0.087 \theta[/tex]

Dividing both sides of the equation by 0.087, we get;

[tex]51.7=\theta[/tex]

Rounding off to the nearest degree, we have;

[tex]52=\theta[/tex]

Thus, the measure of the minor arc AB is 52°

Answer:

52°

Step-by-step explanation:

Arc length = (theta/360) × 2pi × r

4.5 = (theta/360) × 2 × 3.14 × 5

theta/360 = 45/314

Theta = 51.59235669

The mean per capita income is 16,44516,445 dollars per annum with a standard deviation of 397397 dollars per annum. What is the probability that the sample mean would differ from the true mean by greater than 3838 dollars if a sample of 208208 persons is randomly selected? Round your answer to four decimal places.

Answers

Final answer:

To calculate the probability that the sample mean would differ from the true mean by greater than $38, if a sample of 208 persons is randomly selected, we need to use the Central Limit Theorem. First, we determine the standard error of the mean (SEM) using the formula SEM = standard deviation / square root of sample size. Then, we calculate the Z-score using the formula Z = (sample mean - true mean) / SEM. Finally, we find the probability associated with the Z-score using a Z-table or calculator.

Explanation:

To calculate the probability that the sample mean would differ from the true mean by greater than $38, if a sample of 208 persons is randomly selected, we need to use the Central Limit Theorem.

According to the Central Limit Theorem, the distribution of sample means will be approximately normal regardless of the shape of the population distribution, as long as the sample size is large enough.

Since the sample size is greater than 30, we can assume that the distribution of sample means will be approximately normal.

To calculate the probability, we first need to determine the standard error of the mean (SEM), which is the standard deviation divided by the square root of the sample size. In this case, the SEM = $397 / √208.

Next, we calculate the Z-score using the formula Z = (sample mean - true mean) / SEM = ($38 - 0) / ($397 / √208). Finally, we can use a Z-table or calculator to find the probability associated with the Z-score.

In this case, it is the probability that Z is greater than the calculated Z-score. Hence, the probability that the sample mean would differ from the true mean by greater than $38 is the probability that Z is greater than the calculated Z-score.

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